Question #52580

1 Answer
Nov 20, 2017

Answer:

#"200 g MgCl"_2#

Explanation:

The first thing that you need to do here is to figure out how many grams of magnesium chloride would be produced by the reaction at #100%# yield.

To do that, use the molar mass of magnesium chloride

#3 color(red)(cancel(color(black)("moles MgCl"_2))) * "95.211 g"/(1color(red)(cancel(color(black)("mole MgCl"_2)))) = "285.63 g"#

This means that at #100%#yierld, you would expect the reaction to produce #"285.63 g"# of magnesium chloride.

However, you know that the reaction has a #73%# yield, which means that for every #"100 g"# of magnesium chloride that the reaction could theoretically produce, you only get #"73 g"#.

You can thus say that your reaction will produce

#285.63 color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory"))) * overbrace(("73 g MgCl"_2 color(white)(.)"produced")/(100color(red)(cancel(color(black)("g MgCl"_2color(white)(.)"in theory")))))^(color(blue)("= 73% yield"))#

# = color(darkgreen)(ul(color(black)("200 g MgCl"_2color(white)(.)"produced")))#

The answer is rounded to one significant figure, the number of sig figs you have for the number of moles of magnesium chloride that would be produced at #100%# yield.